I've got three maths question...? & i need solution of all...!! plz...!?

Hello.

I don't know how to answer your question, because these are just expressions, not equations. There's nothing to solve.

a^6 - 9a^3 + 8 = [a^3]^2 - 9[a^3] + 8

That's a quadratic in the "variable" [a^3]. Factor to get:

= ([a^3] - 1)([a^3] - 8)

That's a difference of cubes in each factor, so use x^3 - y^3 = (x - y)(x^2 + xy + y^2)

= (a^3 - 1^3)(a^3 - 2^3)

= [(a - 1)(a^2 + a + 1)] [(a - 2)(a^2 + 2x + 4)

Those quadratic terms don't factor (use the discriminant test) so that's it. Simplify to taste.

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8x^6 + 7x^3 - 1 ... do the same thing with [x^3]

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x^3 - x - 2y + 8y^3 .... This is tricker.

The easiest way I see is to factor just the sum of cubes and get lucky.

x^3 + 8y^3 = x^3 + (2y)^3 = (x + 2y)(x^2 - 2xy + 4y^2) .... applying the sum of cubes formula

x^3 + 8y^3 - x - 2y = (x + 2y)(x^2 - 2xy + 4y^2) - x - 2y .... add -x - 2y to both sides

= (x + 2y)(x^2 - 2xy + 4y^2) - (x + 2y) .... factor out -1 in the two right terms

This is the lucky part. There's a common factor of (x + 2y) to factor out:

= (x + 2y)(x^2 - 2xy + 4y^2 - 1)

I think that's as good as you get.

you cannot solve an expression. It must be an equation or an inequality

i will assume though that the questions are = 0

a^6 - 9a^3 + 8 = 0

let a^3 = b

b^2 - 9b + 8 = 0

find two numbers that multiplies to give 8 and adds to give -9

the numbers are -1 and -8

(b -1) ( b -8)

b = 1 or b = 8

but we know that b = a^3

a^3 = 1 and a^ 3 = 8

solving

a = 1 or a = 2

a^6 - 9a^3 + 8 =

(a^3 - 8)(a^3 - 1) =

(a - 2)(a^2 + 2a + 4)(a - 1)(a^2 + a + 1)

8x^6 + 7x^3 - 1 =

(8x^3 - 1)(x^3 + 1) =

(2x - 1)(4x^2 + 2x + 1)(x + 1)(x^2 - x + 1)

x^3 - x - 2y + 8y^3 =

x(x^2 - 1) - 2y(1 - y^2) =

x(x^2 - 1) + 2y(4y^2 - 1) =

x(x + 1)(x - 1) + 2y(2y + 1)(2y - 1)